$\DeclareMathOperator\Cl{Cl}$Let $\Cl(V)$ be the Clifford algebra of a quadratic vector space $(V,Q)$ over $k$, and let $M$ be a (left) $\Cl(V)$-module. The dual Clifford module is typically defined as the linear dual $M^\ast=\operatorname{Hom}_k(M,k)$ with the Clifford action given by defining $g\cdot \phi$ as $g\cdot \phi\colon m\mapsto \phi(g^{\perp}m)$, where $\perp$ is the standard antiinvolution of $\Cl(V)$ given by $(v_1v_2\dotsm v_n)^\perp=v_n\dotsm v_2v_1$. This is similar (not to say it is the same) to the dual representation of a finite group $G$, where the standard antiinvolution of $kG$ is $g\mapsto g^{-1}$.

Yet there is another, a priori distinct and actually more natural, definition of a dual Clifford module: one can consider ${M^\vee}=\operatorname{Hom}_{\Cl(V)}(M,\Cl(V))$ with its natural right $\Cl(V)$-module structure.

This is equivalently a left $\Cl(V)^\text{opp}$-module structure, and since $\perp$ is an algebra isomorphism from $\Cl(V)$ to $\Cl(V)^\text{opp}$, we see that $M^\vee=\operatorname{Hom}_{\Cl(V)}(M,\Cl(V))$ carries a left $\Cl(V)$-module structure via $\perp$.

Now, one may wonder whether $M^\ast$ and $M^\vee$ are isomorphic as left $\Cl(V)$-modules. Equivalently, and more naturally, one is wondering whether $M^\ast$ and $M^\vee$ are isomorphic as right $\Cl(V)$-modules, where the right $\Cl(V)$-module structure on $M^\ast$ is given by $\phi\cdot g\colon m\mapsto \phi(gm)$.

For the representations of a finite group $G$ this is precisely what happens: $M^\ast$ and $M^\vee=\operatorname{Hom}_{kG}(M,kG)$ are isomorphic as right $kG$-modules. An explicit isomorphism $M^\ast\to M^\vee$ is given by

$$\phi\mapsto\left(m \mapsto \sum_{g\in G} \phi(g^{-1}m)g\right).$$

The inverse isomorphism is simply picking the coefficient of the unit element of $G$.

For Clifford algebras things should go the same way, at least over $\mathbb{R}$ and $\mathbb{C}$. Let $(e_1,\dotsc, e_n)$ be an orthonormal basis of $V$, and let $G$ be the subgroup of $Pin(V)$ consisting of the elements $\{\pm e_1^{\epsilon_1}\dotsm e_n^{\epsilon_n}\}$ where $\epsilon_i\in \{0,1\}$. Then $\Cl(V)$-modules are equivalently $G$-modules such that the element $-1$ of $G$ acts as the multiplication by $-1$. One can then define an isomorphism $M^\ast\to M^\vee$ by

$$\phi\mapsto\left(m \mapsto \frac{1}{2}\sum_{g\in G} \phi(g^{-1}m)g\right)=\left(m \mapsto \sum_{g\in G^+} \phi(g^{-1}m)g\right),$$

where $G^+\subset G$ is the subset $\{e_1^{\epsilon_1}\dotsm e_n^{\epsilon_n}\}$ of “positive” elements. Here the sums are done in $\Cl(V)$. To write the inverse isomorphism one notices that there exist a linear morphism $\pi\colon \Cl(V)\to k$ that sends $1$ to $1$ and all other elements of $G^+$ to $0$.

Now the questions (assuming all of the above is correct), are:

i) can one give the isomorphism of left Clifford modules $M^\ast\cong M^\vee$ in an intrinsic way, without using an orthonormal basis (this would possibly give an isomorphism over an arbitrary field $k$)?

ii) if the intrinsic approach fails, can one at least show that the isomorphism above is independent of the chosen orthonormal basis? (This should be a simple check, but I'm postponing it in the hope of a positive answer to i).)


2 Answers 2


$ \newcommand\Cl{\mathrm{Cl} \newcommand\tr{\mathop{\mathrm{tr}}}} \newcommand\Ext{{\textstyle\bigwedge}} \newcommand\form[1]{\langle#1\rangle} \newcommand\Hom{\mathop{\mathrm{Hom}}} \newcommand\rev[1]{#1^\perp} \newcommand\doub\mathfrak $Suppose $k$ has characteristic $\ne 2$ and that $V$ is finite dimensional (which I think you assumed implicitly). Let $V^*$ be the dual of $V$. We will also need to assume that $Q$ is nondegerate, i.e. its bilinear form gives an isomorphism $V \cong V^*$.

Then $\dim\Cl(V, Q) = 2^n$; if we define the trace of an element of $\Cl(V, Q)$ as the trace of its left multiplication $$ \tr X = \tr(Y \mapsto XY) $$ then your $\pi : \Cl(V, Q) \to k$ is exactly $$ \pi(X) = \frac1{2^n}\tr X. $$ This can be confirmed by using an orthonormal basis. Alternatively, we could construct the canonical isomorphism $\Cl(V, Q) \cong \Ext V$ where then $$ \pi(X) = \form{X}_0 $$ where the RHS is the projection onto the scalar (i.e. grade 0) part imported from $\Ext V$ into $\Cl(V,Q)$. Briefly, we can do this by defining $\wedge : V \times \Cl(V,Q) \to \Cl(V,Q)$ by $v\wedge X = \frac12(vX + \hat Xv)$ where $\hat X$ is the main involution applied to $X$ (i.e. the involution that negates all vectors); considering the map $V \to \Hom_k\Cl(V,Q)$ given by $v \mapsto v\wedge\cdot$, the universal property of $\Ext V$ extends this to an algebra homomorphism $\varphi : \Ext V \to \Hom_k\Cl(V, Q)$, and the map $X \mapsto \varphi(X)(1)$ can be shown to be the desired isomorphism. This is a particular instance of Chevalley's maps $\Cl(V, Q) \cong \Cl(V, Q+Q')$; see the notes The Clifford algebra and the Chevalley map — a computational approach by Darij Grinberg for a general approach descending from the tensor algebra.

The bilinear form associated to $Q$ $$ B(v, w) = \frac12(Q(v + w) - Q(v) - Q(w)) $$ induces an isomorphism $\flat : V \to V^*$; using this to apply $Q$ to $V^*$, the universal properties of the Clifford algebras $\Cl(V, Q), \Cl(V^*, Q)$ extend $\flat$ to an algebra isomorphism $\Cl(V, Q) \to \Cl(V^*, Q)$. Using the natural bilinear pairing on $\Ext V^*\times\Ext V \to k$ $$ (v^*_l\wedge v^*_{l-1}\wedge\dotsb\wedge v^*_1,\; v_1\wedge v_2\wedge\dotsb\wedge v_m) \mapsto \delta_{lm}\det\bigl(v^*_i(v_j)\bigr)_{i,j=1}^m $$ then induces a linear isomorphism $\Ext V^* \to (\Ext V)^*$, and composing with $\flat$ finally gives a linear isomorphism $\flat' : \Cl(V, Q) \to \Cl(V, Q)^*$; the bilinear form associated to this turns out to be exactly $$ (X, Y) \mapsto \form{XY}_0. $$ We can actually get the pairing $\Ext V^*\times\Ext V \to k$ by considering the natural Clifford algebra on $V^*\oplus V$ and using the trace definition of the scalar part, but I will not digress further.

Now we define the isomorphism $M^\vee \to M^*$ by $$ \psi \mapsto \Phi_\psi,\quad \Phi_\psi(m) = \form{\psi(m)}_0. $$ This is a homomorphism since $\form{XY}_0 = \form{YX}_0$ and so $$ \Phi_{\psi\cdot g}(m) = \form{\psi(m)g}_0 = \form{g\psi(m)}_0 = \form{\psi(gm)}_0 = (\Phi_\psi\cdot g)(m). $$ The inverse is given by $$ \phi \mapsto \Psi_\phi,\quad \Psi_\phi(m) = (g \mapsto \phi(gm))^{\sharp'} $$ where $\sharp' = (\flat')^{-1}$. Observe: $$ \Psi_{\Phi_\psi}(m) = (g \mapsto \Phi_\psi(gm))^{\sharp'} = (g \mapsto \form{\psi(gm)}_0)^{\sharp'} = (g \mapsto \form{g\psi(m)}_0)^{\sharp'} = \psi(m). $$

Your formula in terms of $G^+$ comes from expressing $\sharp'$ using that basis; the reciprocal basis of $G^+$ is $\{g^{-1} \;:\; g \in G^+\}$.

I have no idea how to handle characteristic 2. The main issue is that there is no canonical isomorphism with $\Ext V$ in this case, and we can't even use the trace since $\tr X = 0$ for all $X$.

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    $\begingroup$ MathJax note: a blank line outside of math mode after command definitions will produce a literal blank line in the rendered source; one must either remove the newline, or put it in math mode (so that in either case the $ that ends the preamble is on the same line as the first line of actual text). I have edited accordingly. $\endgroup$
    – LSpice
    Nov 24, 2022 at 1:40

$\DeclareMathOperator\Cl{Cl}\DeclareMathOperator\char{char}$Not sure about i), but at least ii) is true, so the isomorphism is canonical (at least over $\mathbb{R}$ or $\mathbb{C}$). To see this it is enough to show that the morphism $\pi\colon \Cl(V)\to k$ mentioned in the question is actually canonical. Indeed, if $\char(k)=0$ there is a canonical linear isomorphism $ \bigwedge^\bullet V\to \mathrm{Cl}(V)$ given by $$ v_1\wedge v_2\wedge\dotsb \wedge v_k\mapsto \frac{1}{k!}\sum_{\sigma\in S_k}(-1)^\sigma v_{\sigma(1)}\cdot v_{\sigma(2)}\dotsm v_{\sigma(k)}. $$
Via this isomophism, $\Cl(V)$ inherits a natural grading (as a vector space, not as an algebra!), whose degree zero component is the 1-dimenional subspace generated by the unit element of $\Cl(V)$. We have therefore a distinguished canonical isomorphism $\Cl(V)^0\cong k$ and the projection on the degree zero component $\pi_0\colon \Cl(V)\to \Cl(V)^0$ is consequently a canonical $k$-linear map $\pi_0\colon \Cl(V)\to k$. It is now a simple check to see that, for any orthogonal basis $(e_1,e_2,\dotsc, e_n)$ of $V$ the linear isomorphism $ \bigwedge^\bullet V\to \Cl(V)$ maps $e_{i_1}\wedge e_{i_2}\wedge \cdots e_{i_k}$ to $e_{i_1}\cdot e_{i_2} \cdots e_{i_k}$, so that independently of the choice an orthonormal basis of all of the elements $g\in G^+$ with $g\neq 1$ have strictly positive degree. The a priori basis dependent map $\pi$ mentioned in the question is therefore nothing but the canonical projection on the the degree zero component $\pi_0$.

(The canonical linear isomorphism $\bigwedge^\bullet V\to \Cl(V)$ actually exists for any characteristic different from 2, but the formula is not as nice as in the characteristic zero case: one shows that the linear isomorphism induced by the basis bijection $e_{i_1}\wedge e_{i_2}\wedge \dotsb \wedge e_{i_k} \leftrightarrow e_{i_1}\cdot e_{i_2} \dotsm e_{i_k}$ is independent of the choice of an orthogonal basis $(e_1,e_2\dotsc, e_n)$)

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    $\begingroup$ You divide by $k!$ so it is a bit strange that you just say that the characteristic is different from $2$ ;-) $\endgroup$ Oct 28, 2022 at 17:23
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    $\begingroup$ Argh, right! Actually it is true that there is a canonical linear isomorphism if char(k) is different from 2: one writes it in the other direction, i.e., $Cl(V)\to \wedge^\bullet V$ using an orthogonal basis and then showing it is actually basis independent; but if char(k)=0 one can more elegantly write it as $\wedge^\bullet V\to Cl(V)$ with no choice at all. Thanks! Correcting now. $\endgroup$ Oct 28, 2022 at 17:56
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    $\begingroup$ See my answer here which constructs the canonical isomorphism ${\bigwedge}V \cong \mathrm{Cl}(V)$ without resorting to orthogonal bases (though I do use them to prove the map is bijective). I also want to point out that $\pi$ is actually just the normalized trace: define $\mathrm{tr}(x) = \mathrm{tr}(y \mapsto xy)$ for $x \in \mathrm{Cl}(V)$. Then $\pi(x) = \tfrac1{2^n}\mathrm{tr}(x)$ where $n$ is the dimension of $V$. $\endgroup$ Nov 7, 2022 at 0:29

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